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Thread starter 14 years ago
#1
Self teaching P6 doesn't seem to be a problem so far but occasionally i encounter a question that I have trouble with. I would appreciate some assistance with the following.

Expand e^(x^2 - x) as a power series, showing the first 3 non zero terms.

Here's my working:
f(x) = e^(x^2 - x)
f'(x) = (2x-1)e^(x^2 - x)
f''(x) = 2e^(x^2 - x) + (2x-1)^2.e^(x^2-x)

f(0) = 1, f'(0) = -1, f''(0) = 3

Hence f(x) = 1 -x + (3/2)x^2.
However, the book gives: 1+x-(x^2 / 2) ..

Any ideas where i've gone wrong?
0
14 years ago
#2
(Original post by Gaz031)
Self teaching P6 doesn't seem to be a problem so far but occasionally i encounter a question that I have trouble with. I would appreciate some assistance with the following.

Expand e^(x^2 - x) as a power series, showing the first 3 non zero terms.

Here's my working:
f(x) = e^(x^2 - x)
f'(x) = (2x-1)e^(x^2 - x)
f''(x) = 2e^(x^2 - x) + (2x-1)^2.e^(x^2-x)

f(0) = 1, f'(0) = -1, f''(0) = 3

Hence f(x) = 1 -x + (3/2)x^2.
However, the book gives: 1+x-(x^2 / 2) ..

Any ideas where i've gone wrong?
There's nothing wrong with your working that I can see.

Another way would be to stick x^2-x into the e^x power series to get

exp(x^2-x) = 1 + (x^2-x) + (x^2-x)^2/2! + ....

= 1 + x^2 - x + x^2/2 + ...

= 1 - x + (3/2)x^2 + ...

Which also gives your answer, so I think the book's wrong.
0
14 years ago
#3
(Original post by Gaz031)
Self teaching P6 doesn't seem to be a problem so far but occasionally i encounter a question that I have trouble with. I would appreciate some assistance with the following.

Expand e^(x^2 - x) as a power series, showing the first 3 non zero terms.

Here's my working:
f(x) = e^(x^2 - x)
f'(x) = (2x-1)e^(x^2 - x)
f''(x) = 2e^(x^2 - x) + (2x-1)^2.e^(x^2-x)

f(0) = 1, f'(0) = -1, f''(0) = 3

Hence f(x) = 1 -x + (3/2)x^2.
However, the book gives: 1+x-(x^2 / 2) ..

Any ideas where i've gone wrong?
is this from the Heinemann P6 text book? if so, which page...

your working seems fine to me, you might want to recheck the second derivitive f''(x), it kinda hard to read the some of the computer script! 0
Thread starter 14 years ago
#4
Page 5 Q15.
I'm just about to start elementary transformations from the z-plane to the w-plane but decided i would go back and get this checked.

Thanks for confirming that i haven't made any mistakes.
0
14 years ago
#5
Yeah i just did it... i also get same answer.
1-x+((3x^2) / 2)
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Thread starter 14 years ago
#6
Here are two questions i was having trouble with:

Find in the form a+ib:

Arcsin2. (4m+1)pi/2 + iarcosh2

Arcsini. npi +iarsinh[(-1)^n]

Explanations as to how to find both the real and imaginary parts would be appreciated.

Also. As for showing how one particle moves as another moves in a certain way due to a transformation: IE: A moving along the negative real axis as B moves along the imaginary axis. Is it enough to show a few corresponding points ?
0
14 years ago
#7
(Original post by Gaz031)
Here are two questions i was having trouble with:

Find in the form a+ib:

Arcsin2 = (4m+1)pi/2 + iarcosh2

Arcsini = npi +iarsinh[(-1)^n]

Explanations as to how to find both the real and imaginary parts would be appreciated.

Also. As for showing how one particle moves as another moves in a certain way due to a transformation: IE: A moving along the negative real axis as B moves along the imaginary axis. Is it enough to show a few corresponding points ?
The last paragraph is too vague for me - don't know what you're expecting.

The first two questions seemed weird to be until I realised they were the question and answer. The dot I took for a multiplication sign.

Can I just explain how to solve

sin(x+iy) = a+ib

in general and then you can apply this to your two questions. Note that the solutions won't be unique as sin (even on the complex numbers) has period 2pi.

sin(x+iy) = sinx cos(iy) + sin(iy) cosx

= sinx coshy + i sinhy cosx

So we need to solve

sinx coshy = a
cosx sinhy = b

You could get x by recalling cosh^2y-sinh^2y = 1 to eliminate y;
and could get y by recalling cos^2x+sin^2x=1 to eliminate x.

So to find arcsin2 we have

sinx coshy = 2
cosx sinhy = 0

From the second cosx = 0 (as y=0 leads to no solns) and so sinx = +/-1, but again sinx =-1 leads to no solutions as coshy>=1.

So sinx = 1 and coshy = 2 which means the solutions are

x = (4m+1)pi/2 and y = arcosh2

(this y has a positive and negative solution).

Argues similarly for the other one.
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Thread starter 14 years ago
#8
Thanks for the help. So i need to seperate the real and imaginary parts then compare 'coefficients'.
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14 years ago
#9
(Original post by Gaz031)
Thanks for the help. So i need to seperate the real and imaginary parts then compare 'coefficients'.
Yes - that's one way. Another way is to write

sinz = [e^(iz)-e^{-iz)]/2i and solve the resulting quadratic in e^iz.
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Thread starter 14 years ago
#10
(Original post by RichE)
Yes - that's one way. Another way is to write

sinz = [e^(iz)-e^{-iz)]/2i and solve the resulting quadratic in e^iz.
I tried that first but had trouble getting the right real and imaginary parts. I think the one you mentioned first seems to be the better one (and the book seems to use it too, as they mention arcosh in an answer).
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